Let $K$ be a compact metric space, $E$ be a separable metric space, $C(K,E)$ denote the space of continuous functions from $K$ to $E$ equipped with the supremum metric and $$\pi_t:C(K,E)\to E\;,\;\;\;x\mapsto x(t)$$ for $\in K$.
Now, let $\mathcal M_1(C(K,E))$ denote the space of probability measures on $\mathcal B(C(K,E))$ equipped with the topology of weak convergence.
Question 1: Let $\mathcal F\subseteq\mathcal M_1(C(K,E))$ and $t\in K$. How can we show that $(\mu\circ\pi_t^{-1})_{m\in\mathcal F}$ is tight, i.e. for all $\varepsilon>0$ there is a compact $K\subseteq E$ with $$\mu\left(\left\{x\in C(K,E):x(t)\in E\setminus C\right\}\right)<\varepsilon\tag1$$ for all $\mu\in\mathcal F$?
Question 2: Let $$\omega_x(\delta):=\sup_{\substack{s,\:t\:\in\:K\\d_K(s,\:t)\:\le\:\delta}}d_E(x(s),x(t))$$ for $x\in C(K,E)$ and $\delta\ge0$. Now let $X_n$ be a $C(K,E)$-valued random variable on a probability space $(\Omega,\mathcal A,\operatorname P)$. Assume $(\operatorname P\circ\:X_n^{-1})_{n\in\mathbb N}$ is tight. It's then easy to show that for all $\varepsilon>0$, there is a compact $B\subseteq C(K,E)$ with $$\limsup_{n\to\infty}\operatorname P\left[X_n\in B^c\right]<\varepsilon\tag2.$$ By the Arzelà-Ascoli theorem, there is a $\delta>0$ with $\omega_x(\delta)<\varepsilon$ for all $x\in B$ and hence $$\limsup_{n\to\infty}\operatorname P\left[\omega_{X_n}(\delta)>\varepsilon\right]<\varepsilon\tag3.$$ How can we conclude $$\lim_{\delta\to0+}\limsup_{n\to\infty}\operatorname E\left[\omega_{X_n}(\delta)\wedge 1\right]=0\tag4?$$
$(4)$ reminds me of the situation where a sequence of real-valued random $(Y_n)_{n\in\mathbb N}$ variables converges in probability (and hence in distribution) to a random variable $Y$. From this we can conclude $\operatorname E\left[Y_n\wedge 1\right]\to\operatorname E\left[Y\wedge 1\right]$, since $y\mapsto y\wedge 1$ is bounded and continuous.
I am afraid that without some strong hypotheses on $\mathcal F$ the answer to question 1 is negative. Let $S$ be any dense subset of $C(K,E)$. For simplicity, take $S = C(K,E)$ on a first reading. For notational simplicity, let me call $p$ what you call $\pi_t$. Let $\mathcal F = \{ \delta_f \mid f \in S\}$, where $\delta_f$ is the Dirac measure at $f \in S$. I claim that $p_* \mathcal F$ is not tight, where $p_*$ is the pushforward, i.e. $p_* \mu = \mu \circ p^{-1}$.
If it were, then for some $\varepsilon <1$ (the strict inequality is essential!) there would exist some compact $C \subseteq E$ such that $(p_* \mu) (E \setminus C) < \varepsilon$ for all $\mu \in \mathcal F$. But this means that $\delta_f (p^{-1} (E \setminus C)) < \varepsilon < 1$ for every $f \in S$ and, since $\delta_f$ takes only the values $0$ and $1$, it means that $\delta_f (p^{-1} (E \setminus C)) = 0$ or, in other words, $f \notin p^{-1} (E \setminus C)$ or, equivalently, $f(t) \notin E \setminus C$ for all $f \in S$. Equivalently, $p(S) \subseteq C$, whence $p(C(K,E)) \subseteq C$, by the density of $S$ in $C(K,E)$ and the continuity of $p$.
But this is nonsense! If $E = \mathbb R$ then clearly $p (C(K,\mathbb R)) = \mathbb R$ (evaluating all the continuous functions on $K$ at some $t \in K$ definitely covers the whole $\mathbb R$), which shows that $p(C(K,E))$ cannot stay inside some compact $C$.
One possible way out of this would probably be to assume $\mathcal F$ tight, in which case there is not much left to prove. Some other possibility might be to assume $E$ compact, but I do not know if this is enough. The negative answer shouldn't surprise you, though: you start with an arbitrary $\mathcal F$ and hope to obtain some compactness of $p_* \mathcal F$. You want to get way too much from way too little!